where in the last line we interchanged the second and the third components for brevity.
We consider the relative commutant of the wedge shifted by \(a = (a_+, a_-), a_+ < 0, a_- > 0\) .
With the above interchanged notation, it is clear that the \(-\) -component (the 3rd and 4th tensor components)
of the intersection is simply \({\mathcal {A}}_0((0,a_-))\otimes {\mathcal {A}}_0((0,a_-))\) .
As for the \(+\) -component (the 1st and 2nd tensor components), one observes that
it is almost same as the intersection calculated in [1]}.
The only change is that the direction of the 2nd component is reversed. This does not affect
the proof, indeed, the inner symmetry commutes with translation and the positivity of energy
is used only through Reeh-Schlieder property hence is not essential.
Thus we have
\(\widetilde{\mathcal {M}}_{\mathrm {c},k}\cap {\hbox{\rm Ad\,}}\widetilde{T}_{\mathrm {c}}(a)(\widetilde{\mathcal {M}}_{\mathrm {c},k}) = \left({\mathcal {A}}_0^{\alpha _0}((a_+,0))\otimes {\mathcal {A}}_0^{\alpha _0}((a_+,0))\right)\otimes \left({\mathcal {A}}_0((0,a_-))\otimes {\mathcal {A}}_0((0,a_-))\right),\) 
